on2t1inf8b

2022-07-20

The original problem is:
If a, b, c, d are the position vectors of points A, B, C, D respectively such that
$\left(\stackrel{\to }{a}-\stackrel{\to }{d}\right).\left(\stackrel{\to }{b}-\stackrel{\to }{c}\right)=\left(\stackrel{\to }{b}-\stackrel{\to }{d}\right).\left(\stackrel{\to }{c}-\stackrel{\to }{a}\right)=0$
then prove that D is the orthocentre of $\mathrm{\Delta }$ ABC.
How do we go about proving that a point is the orthocentre of a triangle? I've tried expanding the dot product but I don't seem to get anywhere.

kuglatid4

Expert

That formula states that $AD\perp BC$ (so that $D$ is on the altitude from $A$ to $BC$) and that $AC\perp BD$ (so that $D$ is on the altitude from $B$ to $AC$). As $D$ is on two altitudes of the triangle, it is its orthocentre.